How do you solve $m^{2} + 8 m + 15 = 0$ $m^2 + 8m + 15 = 0$ by completing the square?

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How do you solve $m^{2} + 8 m + 15 = 0$ $m^2 + 8m + 15 = 0$ by completing the square?

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Answer 1 · 2016-04-02T06:43:34

Complete the square to find $m = - 3$ $m = -3$ or $m = - 5$ $m = -5$

Explanation:

Note that:

${(m + 4)}^{2} = m^{2} + 2 (m) (4) + 4^{2} = m^{2} + 8 m + 16$ $(m+4)^2 = m^2+2(m)(4) + 4^2 = m^2+8m+16$

So add $1$ $1$ ti both sides of the equation to get:

$m^{2} + 8 m + 16 = 1$ $m^2+8m+16 = 1$

which we can write as:

${(m + 4)}^{2} = 1$ $(m+4)^2 = 1$

Then take the square root of both sides, allowing for both positive and negative square roots to find:

$m + 4 = \pm \sqrt{1} = \pm 1$ $m+4 = +-sqrt(1) = +-1$

Subtract $4$ $4$ from both sides to get:

$m = - 4 \pm 1$ $m = -4+-1$

That is $m = - 5$ $m = -5$ or $m = - 3$ $m = -3$

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Alternative method

Use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = (m + 4)$ $a = (m+4)$ and $b = 1$ $b=1$ as follows:

$0 = m^{2} + 8 m + 15$ $0 = m^2+8m+15$

$= {(m + 4)}^{2} - 16 + 15$ $= (m+4)^2-16+15$

$= {(m + 4)}^{2} - 1$ $= (m+4)^2-1$

$= {(m + 4)}^{2} - 1^{2}$ $= (m+4)^2 - 1^2$

$= ((m + 4) - 1) ((m + 4) + 1)$ $= ((m+4) - 1)((m+4) + 1)$

$= (m + 3) (m + 5)$ $= (m+3)(m+5)$

So $m = - 3$ $m = -3$ or $m = - 5$ $m = -5$

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How do you solve $m^{2} + 8 m + 15 = 0$ $m^2 + 8m + 15 = 0$ by completing the square?

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How do you solve $m^{2} + 8 m + 15 = 0$ $m^2 + 8m + 15 = 0$ by completing the square?